Problem: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 2} = \dfrac{81}{x - 2}$
Multiply both sides by $x - 2$ $ \dfrac{x^2}{x - 2} (x - 2) = \dfrac{81}{x - 2} (x - 2)$ $ x^2 = 81$ Subtract $81$ from both sides: $ x^2 - (81) = 81 - (81)$ $ x^2 - 81 = 0$ Factor the expression: $ (x + 9)(x - 9) = 0$ Therefore $x = -9$ or $x = 9$ The original expression is defined at $x = -9$ and $x = 9$, so there are no extraneous solutions.